In Silicon Valley, a number of computer firms were found to be contaminating underground water supplies with toxic chemicals stored in leaking underground containers. A water quality control agency ordered the companies to take immediate corrective action and contribute to a monetary pool for the testing and cleanup of the underground contamination. Suppose that the required monetary pool (in millions of dollars) is given by the following function, where $$ x $$ is the percentage (expressed as a decimal fraction) of the total contaminant removed. Complete parts (A) through (E). $$ P(x)=\frac{11 x}{1-x} \quad 0 \leq x

Question

In Silicon Valley, a number of computer firms were found to be contaminating underground water supplies with toxic chemicals stored in leaking underground containers. A water quality control agency ordered the companies to take immediate corrective action and contribute to a monetary pool for the testing and cleanup of the underground contamination. Suppose that the required monetary pool (in millions of dollars) is given by the following function, where $$ x $$ is the percentage (expressed as a decimal fraction) of the total contaminant removed. Complete parts (A) through (E). $$ P(x)=\frac{11 x}{1-x} \quad 0 \leq x<1 $$ (A) Where is $$ P(x) $$ increasing? Decreasing? Select the correct choice below and fill in the answer box(es) to complete your choice. (Type your answer in interval notation. Use integers or fractions for any numbers in the expression. Use a comma to separate answers as needed.) A. $$ P(x) $$ is increasing on $$ \square $$ and decreasing on B. $$ P(x) $$ is increasing on C. $$ P(x) $$ is decreasing on $$ \square $$.

UpStudy AI · Accepted Answer

**Step 1. Compute the Derivative**

We start with the function  
$$
P(x)=\frac{11x}{1-x}.
$$

To determine where the function is increasing or decreasing, we first compute its derivative using the quotient rule. The quotient rule states that if  
$$
f(x)=\frac{u(x)}{v(x)},
$$
then  
$$
f'(x)=\frac{u'(x)v(x)-u(x)v'(x)}{[v(x)]^2}.
$$

For our function, let  
$$
u(x)=11x \quad \text{and} \quad v(x)=1-x.
$$

We have  
$$
u'(x)=11 \quad \text{and} \quad v'(x)=-1.
$$

Thus, applying the quotient rule:
$$
P'(x)=\frac{11(1-x) - 11x(-1)}{(1-x)^2}.
$$

**Step 2. Simplify the Derivative**

Simplify the numerator:
$$
11(1-x) - 11x(-1) = 11(1-x) + 11x = 11(1-x+x) = 11.
$$

Thus, the derivative is:
$$
P'(x)=\frac{11}{(1-x)^2}.
$$

**Step 3. Determine Where $$ P(x) $$ is Increasing or Decreasing**

The derivative,  
$$
P'(x)=\frac{11}{(1-x)^2},
$$
has a constant numerator $$11$$ (which is positive) and a denominator $$(1-x)^2$$, which is always positive for all $$ x $$ in the domain $$[0,1)$$. Therefore, the derivative is positive:
$$
P'(x) > 0 \quad \text{for all} \quad 0 \le x < 1.
$$

Since the derivative is positive throughout the domain, the function is increasing on the entire interval $$[0,1)$$ and is never decreasing.

**Final Answer**

B. $$ P(x) $$ is increasing on $$[0,1)$$.
$$ P(x) $$ is increasing on $$[0,1)$$.

Solución de inteligencia artificial de Upstudy

Responder

Solución

The Deep Dive

preguntas relacionadas

Latest Calculus Questions